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halloweenjack
Dec15-08, 01:00 AM
So this may be an easy question, I'm not sure, but I could not find an answer online. Basically here is the scenario.

There are two balls, one has a mass of 1g and the other has a mass of 100g. The 1g ball is traveling at 10m/s, and strikes the other ball. According to momentum, m1v1=m2v2. Using this equation, we have 1g(10m/s)=100g(v2). v2 then equals 0.1 m/s.

The problem I have is that when using kinetic energy, I get a different velocity of the second ball. The first ball has an energy of .5(1g)(10m/s)^2, which is 50 joules. If the first ball then transfers these 50 joules to the second ball, using the same equation I get the second ball having a final velocity of 50=.5(100g)(v^2), or 1 m/s.

Why am I getting different answers? Shouldn't both equations give the same answer?

D H
Dec15-08, 01:13 AM
The answer is simple: The collision as you have established it is not elastic. In other words, energy isn't conserved. In an elastic, head-on collision the small ball will bounce back with nearly the same speed it had before the collision and the big ball will move forward with a tiny velocity:

m_1 v_{1,\text{init}} = m_1 v_{1,\text{final}} + m_2v_2

and

m_1 v_{1,\text{init}}^2 = m_1 v_{1,\text{final}}^2 + m_2v_2^2

Kate21
Dec15-08, 05:21 PM
Thanks for you help. This helps me in some practice I am getting in to be better in mathematics.